Zajj Daugherty...2018/07/16 · C[[x 1;x 2;:::]] consisting of series where the coe cients on the...

Quasisymmetric power sums

Zajj Daugherty

Joint work withCristina Ballantine, Angela Hicks,Sarah Mason, and Elizabeth Niese

Some combinatorics

Partitions:

= (5, 4, 4, 2) = λ

Compositions:

= (4, 2, 5, 4) = α

For a composition α,

let |α| be the size (# boxes) of α;

let `(α) be the length (# parts) of α; and

let α̃ be the rearrangement of the parts of α into decreasing order.

For example, |α| = 15, `(α) = 4, and α̃ = λ.

For compositions α and β, we say α refines β, written α 4 β, if β canbe built by combining adjacent parts of α. For example,

4 , 64

Some combinatorics

Partitions:

= (5, 4, 4, 2) = λ

Compositions:

= (4, 2, 5, 4) = α

4 , 64

Some combinatorics

Partitions:

= (5, 4, 4, 2) = λ

Compositions:

= (4, 2, 5, 4) = α

4 , 64

Let Sym be the ring of symmetric functions.

Many favorite bases indexed by integer partitions λ ` n:• Monomial symmetric functions

mλ =∑α̃=λ

i1<i2<···<i`

xα1i1xα2i2· · ·xα`i`

• Complete homogeneous symmetric functions

hr =∑|α|=r

i1<i2<···<i`

xα1i1xα2i2· · ·xα`i` =

∑|λ|=r

mλ, hλ = hλ1hλ2 · · · .

• Elementary symmetric functions

er =∑

1≤i1<i2<···<ir

xi1 · · ·xir = m(1,1,...,1) eλ = eλ1eλ2 · · · .

? Power sum symmetric functions

pr = xr1 + xr2 + · · · = m(r), pλ = pλ1pλ2 · · ·

We have a scalar product 〈, 〉 : Sym⊗ Sym→ C defined by

〈hλ,mµ〉 = δλ,µ,

so that the homogeneous and monomial functions are dual.

Underthis pairing, we have

〈pλ, pµ〉 = zλδλµ,

where zλ is the size of the stabilizer of a permutation of cycle type λ:

zλ =∏k

ak!kak , ak = #{pts of length k} Ex: z = 2! 32.

Generating functions:

H(t) =∑k≥0

hktk =

∏i≥1

(1− xit)−1

E(t) =∑k≥0

ektk =

∏i≥1

(1 + xit)

Note H(t) = 1/E(−t).

P (t) =∑k≥0

pktk =

dtln(H(t)) =

dtln(1/E(−t))

so that the homogeneous and monomial functions are dual. Underthis pairing, we have

zλ =∏k

H(t) =∑k≥0

hktk =

∏i≥1

(1− xit)−1

E(t) =∑k≥0

ektk =

∏i≥1

(1 + xit)

P (t) =∑k≥0

pktk =

dtln(H(t)) =

dtln(1/E(−t))

so that the homogeneous and monomial functions are dual. Underthis pairing, we have

zλ =∏k

H(t) =∑k≥0

hktk =

∏i≥1

(1− xit)−1

E(t) =∑k≥0

ektk =

∏i≥1

(1 + xit)

P (t) =∑k≥0

pktk =

dtln(H(t)) =

dtln(1/E(−t))

The ring of noncommutative symmetric functions NSym is theC-algebra generated freely by e1, e2, . . . .

Analogous bases indexed by compositions α.

• Noncommutative elementary: eα = eα1 · · · eα` . Ab(eα) = eα̃

• Noncom. homog.: hα = hα1 · · ·hα` , where hi is defined by. . .

if E(t) =∑k≥0

ektk and H(t) =

∑k≥0

then H(t) = 1/E(−t). (Recall: H(t) = 1/E(−t) in Sym).Ab(hα) = hα̃

? Noncommutative power sums: two choices, ψ and φ!

In Sym: In NSym:

Type 1: P (t) = ddt ln(H(t)) d

dtH(t) = H(t)Ψ(t)

Type 2: H(t) = exp(∫P (t)dt

)H(t) = exp

(∫Φ(t)dt

)Not the same! (No unique notion of log derivative for power serieswith noncommutative coefficients.) But

Ab(ψα) = pα̃ = Ab(φα)

• Noncommutative elementary: eα = eα1 · · · eα` .

Ab(eα) = eα̃

if E(t) =∑k≥0

ektk and H(t) =

∑k≥0

In Sym: In NSym:

dtH(t) = H(t)Ψ(t)

)H(t) = exp

(∫Φ(t)dt

if E(t) =∑k≥0

ektk and H(t) =

∑k≥0

In Sym: In NSym:

dtH(t) = H(t)Ψ(t)

)H(t) = exp

(∫Φ(t)dt

if E(t) =∑k≥0

ektk and H(t) =

∑k≥0

In Sym: In NSym:

dtH(t) = H(t)Ψ(t)

)H(t) = exp

(∫Φ(t)dt

if E(t) =∑k≥0

ektk and H(t) =

∑k≥0

In Sym: In NSym:

dtH(t) = H(t)Ψ(t)

)H(t) = exp

(∫Φ(t)dt

if E(t) =∑k≥0

ektk and H(t) =

∑k≥0

In Sym: In NSym:

dtH(t) = H(t)Ψ(t)

)H(t) = exp

(∫Φ(t)dt

The ring of quasisymmetric functions QSym is a subring ofC[[x1, x2, . . . ]] consisting of series where the coefficients on themonomials

xα11 xα2

2 · · ·xα`` and xα1

i1xα2i2· · ·xα`i`

are the same, for all i1 < i2 < · · · < i`. In particular, Sym ⊂ QSym.

For example,∑i<j

xix2j = x1x

22 + x1x

23 + x2x

23 + · · ·

is quasisymmetric but not symmetric (the coef. on x21x2 is 0).

Bases of QSym are also indexed by compositions. Namely, themonomial basis has a natural analog:

Mα =∑

i1<i2<···<i`(α)

xα1i1xα2i2· · ·xα`i` , so that mλ =

∑α̃=λ

xα11 xα2

2 · · ·xα`` and xα1

For example,∑i<j

xix2j = x1x

22 + x1x

23 + x2x

23 + · · ·

Mα =∑

i1<i2<···<i`(α)

∑α̃=λ

xα11 xα2

2 · · ·xα`` and xα1

For example,∑i<j

xix2j = x1x

22 + x1x

23 + x2x

23 + · · ·

Bases of QSym are also indexed by compositions.

Namely, themonomial basis has a natural analog:

Mα =∑

i1<i2<···<i`(α)

∑α̃=λ

xα11 xα2

2 · · ·xα`` and xα1

For example,∑i<j

xix2j = x1x

22 + x1x

23 + x2x

23 + · · ·

Mα =∑

i1<i2<···<i`(α)

∑α̃=λ

xα11 xα2

2 · · ·xα`` and xα1

For example,∑i<j

xix2j = x1x

22 + x1x

23 + x2x

23 + · · · = M

Mα =∑

i1<i2<···<i`(α)

∑α̃=λ

The rings NSym and QSym have Hopf algebra structures that aredual to each other under a pairing

〈, 〉 : NSym⊗QSym→ Canalogous to the pairing in Sym.

Namely

〈hλ,mµ〉 = δλ,µ in Sym⊗ Sym,

〈hα,Mβ〉 = δα,β in NSym⊗QSym.

Recall in Sym, the power sum basis is (essentially) self-dual:

〈pλ, pµ〉 = zλδλµ

Question: What is dual to ψ in QSym? to φ?

〈, 〉 : NSym⊗QSym→ Canalogous to the pairing in Sym. Namely

Type 1In Sym the power sum basis is (essentially) self-dual:

In NSym, the type 1 power sum basis ψ is defined by the generatingfunction relation

dtH(t) = H(t)Ψ(t).

This is equivalent to

hα =∑β4α

π(β, α)ψβ,

where π(β, α) is a combinatorial statistic on the refinement β 4 α.So, the dual in QSym will satisfy

ψ∗α =∑β<α

π(α, β)Mβ.

DefineΨα = zα̃ψ

∗α, so that 〈ψα,Ψβ〉 = zα̃δαβ.

dtH(t) = H(t)Ψ(t).

hα =∑β4α

π(β, α)ψβ,

where π(β, α) is a combinatorial statistic on the refinement β 4 α.

So, the dual in QSym will satisfy

ψ∗α =∑β<α

π(α, β)Mβ.

dtH(t) = H(t)Ψ(t).

hα =∑β4α

π(β, α)ψβ,

ψ∗α =∑β<α

π(α, β)Mβ.

dtH(t) = H(t)Ψ(t).

hα =∑β4α

π(β, α)ψβ,

ψ∗α =∑β<α

π(α, β)Mβ.

Computing coefficientsΨα = zα̃

∑β<α

π(α, β)Mβ.

For example, we saw that

First, for each block, we compute the product of the partial sums:

( )=∣∣ ∣∣ · ∣∣∣ ∣∣∣ · ∣∣∣∣ ∣∣∣∣ = 1 · 3 · 4

Then, for α refining β, the coefficient of Mβ in ψ∗α is 1/π(α, β), where

( )π( )

= (1 · 3 · 4)(2)(5)(1 · 2 · 4)

∑β<α

π(α, β)Mβ.

( )=∣∣ ∣∣ · ∣∣∣ ∣∣∣ · ∣∣∣∣ ∣∣∣∣ = 1 · 3 · 4

( )π( )

= (1 · 3 · 4)(2)(5)(1 · 2 · 4)

∑β<α

π(α, β)Mβ.

( )=∣∣ ∣∣ · ∣∣∣ ∣∣∣ · ∣∣∣∣ ∣∣∣∣ = 1 · 3 · 4

( )π( )

= (1 · 3 · 4)(2)(5)(1 · 2 · 4)

∑β<α

π(α, β)Mβ.

( )=∣∣ ∣∣ · ∣∣∣ ∣∣∣ · ∣∣∣∣ ∣∣∣∣ = 1 · 3 · 4

( )π( )

= (1 · 3 · 4)(2)(5)(1 · 2 · 4)

Computing coefficientsFirst, for each block, we compute the product of the partial sums:

=∣∣ ∣∣ · ∣∣∣ ∣∣∣ · ∣∣∣ ∣∣∣ = 1 · 3 · 4

= π( )

As another example, z = 2,

Ψ = z ψ∗ = 2

Ψ + Ψ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

Type 1 QSym power sums refine Sym power sums:

pλ =∑α̃=λ

Computing coefficientsFirst, for each block, we compute the product of the partial sums:

=∣∣ ∣∣ · ∣∣∣ ∣∣∣ · ∣∣∣ ∣∣∣ = 1 · 3 · 4

= π( )

Ψ = z ψ∗ = 2

Ψ + Ψ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

Computing coefficientsAs another example, z = 2,

Ψ = z ψ∗ = 2

Ψ + Ψ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

Ψ = z ψ∗ = 2

Ψ + Ψ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

Ψ = z ψ∗ = 2

Ψ + Ψ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

Theorem: pλ =∑α̃=λ

Ψα, where Ψα = zα̃∑α4β

π(α, β)Mβ.

Proof outline: For compositions α and β, define Oα,β be the set ofordered set partitions (B1, · · · , B`(β)) of {1, · · · , `(α)} satisfying

βj =∑i∈Bj

αi for 1 ≤ j ≤ `(β).

For example, if

α = and β = ,

then Oα,β contains ({1, 3}, {2}) and ({2}, {1, 3}).

It has been shown that

pλ =∑

part’n µ

|Oλ,µ|mµ, so that pλ =∑

comp β

|Oλ,β|Mβ.

π(α, β)Mβ.

βj =∑i∈Bj

αi for 1 ≤ j ≤ `(β).

For example, if

α = and β = ,

pλ =∑

part’n µ

comp β

|Oλ,β|Mβ.

π(α, β)Mβ.

βj =∑i∈Bj

αi for 1 ≤ j ≤ `(β).

For example, if

α = and β = ,

pλ =∑

part’n µ

comp β

|Oλ,β|Mβ.

π(α, β)Mβ.

βj =∑i∈Bj

αi for 1 ≤ j ≤ `(β).

pλ =∑

part’n µ

comp β

|Oλ,β|Mβ.

We combinatorially prove, for a fixed partition λ with size n, and afixed composition β, that

|Oλβ| · |Sλn | =

|Oλβ|n!

zλ=∑α4βα̃=λ

π(α, β),

where Sλn = {σ ∈ Sn of cycle type λ}.

π(α, β)Mβ.

βj =∑i∈Bj

αi for 1 ≤ j ≤ `(β).

pλ =∑

part’n µ

comp β

|Oλ,β|Mβ.

We combinatorially prove, for a fixed partition λ with size n, and afixed composition β, that

|Oλβ| · |Sλn | = |Oλβ|n!

zλ=∑α4βα̃=λ

π(α, β),

where Sλn = {σ ∈ Sn of cycle type λ}.

Two ways of thinking about permutations:I In one-line notation:

σ = 571423689

is the permutation sending

1 7→ 5, 2 7→ 7, 3 7→ 1, and so on. . .

I In cycle notation:

σ = (152763)(4)(8)(9).

Several equivalent ways to write in cycle notation. We say σ iswritten in standard form if

largest element of each cycle is last, andcycles ordered increasingly according to largest element

Ex: (4)(631527)(8)(9)I Let α 4 β of size n, and let σ ∈ Sn. We say σ is consistent withα 4 β if. . .

Ex: let α = (1, 1, 2, 1, 3, 1) and β = (2, 2, 5)Start in one-line notation: 571423689Split according to β: 57||14||23689Add parentheses according to α: (5)(7)||(14)||(2)(368)(9)

If the permutations in each partition are in standard form, then σis consistent.Non-example: 571428369 → (5)(7)||(14)||(2)(836)(9)

σ = 571423689

1 7→ 5, 2 7→ 7, 3 7→ 1, and so on. . .I In cycle notation:

σ = (152763)(4)(8)(9).

σ = 571423689

σ = (152763)(4)(8)(9).

Ex: (4)(631527)(8)(9)

I Let α 4 β of size n, and let σ ∈ Sn. We say σ is consistent withα 4 β if. . .

σ = 571423689

σ = (152763)(4)(8)(9).

σ = 571423689

σ = (152763)(4)(8)(9).

Ex: let α = (1, 1, 2, 1, 3, 1) and β = (2, 2, 5)

Start in one-line notation: 571423689Split according to β: 57||14||23689Add parentheses according to α: (5)(7)||(14)||(2)(368)(9)

σ = 571423689

σ = (152763)(4)(8)(9).

Ex: let α = (1, 1, 2, 1, 3, 1) and β = (2, 2, 5)Start in one-line notation: 571423689

Split according to β: 57||14||23689Add parentheses according to α: (5)(7)||(14)||(2)(368)(9)

σ = (152763)(4)(8)(9).

Ex: (4)(631527)(8)(9)

Ex: let α = (1, 1, 2, 1, 3, 1) and β = (2, 2, 5)Start in one-line notation: 571423689Split according to β: 57||14||23689

Add parentheses according to α: (5)(7)||(14)||(2)(368)(9)

σ = (152763)(4)(8)(9).

Ex: (4)(631527)(8)(9)

σ = (152763)(4)(8)(9).

Ex: (4)(631527)(8)(9)

If the permutations in each partition are in standard form, then σis consistent.

Non-example: 571428369 → (5)(7)||(14)||(2)(836)(9)

σ = (152763)(4)(8)(9).

Ex: (4)(631527)(8)(9)

Cons(1,2,1)4(1,2,1) = {1234, 1243, 1342, 2134, 2143, 2341, 3124,

3142, 3241, 4123, 4132, 4231},

π((1, 2, 1), (1, 2, 1)) = 2

Cons(1,2,1)4(1,3) = {1234, 2134, 3124, 4123},

π((1, 2, 1), (1, 3)) = 2 · 3

Cons(1,2,1)4(3,1) = {1234, 1243, 1342, 2134, 2143, 2341, 3142, 3241},

π((1, 2, 1), (3, 1)) = 1 · 3

Cons(1,2,1)4(4) = {1234, 2134}

π((1, 2, 1), (4)) = 1 · 3 · 4

LemmaFix α 4 β of size n Then

n! = |Consα4β| · π(α, β).

Cons(1,2,1)4(1,2,1) = {1234, 1243, 1342, 2134, 2143, 2341, 3124,

3142, 3241, 4123, 4132, 4231},

π((1, 2, 1), (1, 2, 1)) = 2

Cons(1,2,1)4(1,3) = {1234, 2134, 3124, 4123},

π((1, 2, 1), (1, 3)) = 2 · 3

Cons(1,2,1)4(3,1) = {1234, 1243, 1342, 2134, 2143, 2341, 3142, 3241},

π((1, 2, 1), (3, 1)) = 1 · 3

Cons(1,2,1)4(4) = {1234, 2134}

π((1, 2, 1), (4)) = 1 · 3 · 4

Cons(1,2,1)4(1,2,1) = {1234, 1243, 1342, 2134, 2143, 2341, 3124,

3142, 3241, 4123, 4132, 4231},

π((1, 2, 1), (1, 2, 1)) = 2

Cons(1,2,1)4(1,3) = {1234, 2134, 3124, 4123},

π((1, 2, 1), (1, 3)) = 2 · 3

Cons(1,2,1)4(3,1) = {1234, 1243, 1342, 2134, 2143, 2341, 3142, 3241},

π((1, 2, 1), (3, 1)) = 1 · 3

Cons(1,2,1)4(4) = {1234, 2134}

π((1, 2, 1), (4)) = 1 · 3 · 4

Proof: Let

Aα4β =

`(β)∏i=1

`(α(i))∏j=1

Z/a(i)j Z

, where a(i)j =

j∑r=1

α(i)r ,

so that |Aα4β| = π(α, β). Then there is a bijection

Sn → Consα4β ×Aα4β.(For each permutation written in one-line notation, “cycle” into consistency

with α 4 β, and record cycles with elements of Aα4β . See slides from

“Discrete Math Day” October 2017.)

|Oα4β| · |Sλn | =∑α4βα̃=λ

|Consα4β|.

(Similar proof.)

Therefore

|Oλβ| · |Sλn | =∑α4βα̃=λ

π(α, β),

so that

pλ =∑

comp β

|Oλ,β|Mβ =∑α̃=λ

π(α, β)Mβ,

as desired.

Proof: Let

Aα4β =

`(β)∏i=1

`(α(i))∏j=1

Z/a(i)j Z

, where a(i)j =

j∑r=1

α(i)r ,

so that |Aα4β| = π(α, β). Then there is a bijection

Sn → Consα4β ×Aα4β.(For each permutation written in one-line notation, “cycle” into consistency

with α 4 β, and record cycles with elements of Aα4β . See slides from

“Discrete Math Day” October 2017.)

|Consα4β|.

(Similar proof.)

Therefore

π(α, β),

so that

pλ =∑

comp β

π(α, β)Mβ,

as desired.

|Consα4β|.

(Similar proof.)

Therefore

π(α, β),

so that

pλ =∑

comp β

π(α, β)Mβ,

as desired.

〈pλ, pµ〉 = zλδλµ.

In NSym, the type 2 power sum basis is defined by the generatingfunction relation

H(t) = exp

(∫Φ(t)dt

hα =∑β4α

sp(β, α)φβ,

where sp(β, α) is a combinatorial statistic on the refinement β 4 α.So, the dual in QSym will satisfy

φ∗α =∑β<α

sp(α, β)Mβ.

DefineΦα = zα̃φ

∗α, so that 〈φα,Φβ〉 = zαδαβ.

H(t) = exp

(∫Φ(t)dt

)This is equivalent to

hα =∑β4α

sp(β, α)φβ,

where sp(β, α) is a combinatorial statistic on the refinement β 4 α.

So, the dual in QSym will satisfy

φ∗α =∑β<α

sp(α, β)Mβ.

H(t) = exp

(∫Φ(t)dt

hα =∑β4α

sp(β, α)φβ,

φ∗α =∑β<α

sp(α, β)Mβ.

H(t) = exp

(∫Φ(t)dt

hα =∑β4α

sp(β, α)φβ,

φ∗α =∑β<α

sp(α, β)Mβ.

Computing coefficientsΦα = zα̃

∑β<α

sp(α, β)Mβ.

First, for each block, we compute sp(γ) = `(γ)!∏k γj :

( )= 3!(1 · 2 · 1)

Then, for α refining β, the coefficient of Mβ in ψ∗α is 1/sp(α, β),where

( )sp( )

= 3!(1 · 2 · 1) · 1!(2) · 1!(5) · 3!(1 · 1 · 2)

∑β<α

sp(α, β)Mβ.

( )= 3!(1 · 2 · 1)

( )sp( )

= 3!(1 · 2 · 1) · 1!(2) · 1!(5) · 3!(1 · 1 · 2)

∑β<α

sp(α, β)Mβ.

= 3!(1 · 2 · 1)

= sp( )

= 3!(1 · 2 · 1) · 1!(2) · 1!(5) · 3!(1 · 1 · 2)

Computing coefficientssp( )

= `(γ)!∏k γj = 3!(1 · 2 · 1)

= sp( )

Φ = z φ∗ = 2

Φ + Φ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

= `(γ)!∏k γj = 3!(1 · 2 · 1)

= sp( )

Φ = z φ∗ = 2

Φ + Φ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

= `(γ)!∏k γj = 3!(1 · 2 · 1)

= sp( )

Φ = z φ∗ = 2

Φ + Φ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

= `(γ)!∏k γj = 3!(1 · 2 · 1)

= sp( )

Φ = z φ∗ = 2

Φ + Φ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

Φ = z φ∗ = 2

Φ + Φ = M +M +M

= m +m = m m = p2p1 = p .

Theorem (BDHMN)

pλ =∑α̃=λ

Other results

We also give

• Transition matrices between type 1 and type 2, and tofundamentals.

• Combinatorial proofs of product formulas for type 1 and 2:

ΨαΨβ =zαzβzα·β

∑γ∈α�β

Ψwd(γ), ΦαΦβ =zαzβzα·β

∑γ∈α�β

Φwd(γ).

• Some comments about plethysm, and other algebraic connections.

See arXiv:1710.11613 for more details.

Thanks!

Zajj Daugherty...2018/07/16 · C[[x 1;x 2;:::]] consisting of series where the coe cients on the...

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